Question 88
If $$a^{2}sec^{2} x-b^{2} tan^{2} x$$=$$c^{2}$$ then the value of $$sec^{2} x+tan^{2} x $$ is equal to ($$ b^{2} \neq a^{2}$$)
Solution
Given that $$a^{2}sec^{2} x-b^{2} tan^{2} x$$=$$c^{2}$$
=> $$a^{2}(1+tan^{2} x)-b^{2} tan^{2} x$$=$$c^{2}$$
=> $$a^2+tan^{2}x(a^2-b^2)=c^{2}$$
=> $$tan^{2}x=\frac{c^2-a^2}{a^2-b^2}$$
Therefore,
$$sec^{2} x+tan^{2} x $$
= $$1+2tan^{2}x$$
=$$1+2(\frac{c^2-a^2}{a^2-b^2})$$
=$$\frac{b^{2}+a^{2}-2C^{2}}{b^{2}-a^{2}}$$
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